Question

\(75%\) of a first-order reaction was completed in \(32\) minutes. How long will it take to undergo \(50%\) completion? Given: \(\log 2 = 0.3010\)

• A. \(8\) min
• B. \(12\) min
• C. \(16\) min
• D. \(24\) min

Hint:

For a first-order reaction, 75% completion corresponds to two half-lives because: \[ 100%\rightarrow50%\rightarrow25% \] Thus, if two half-lives take 32 min, one half-life takes 16 min.

Answer 1

The Correct Option is C

Step 1: Interpret 75% completion.
If 75% reaction is completed, then 25% reactant remains. \[ \frac{[R]}{[R]_0}=0.25 \] \[ \frac{[R]_0}{[R]}=4 \]

Step 2: Apply first-order equation.
\[ 32 = \frac{2.303}{k}\log 4 \] Since \[ \log4=2\log2 \] \[ =2(0.3010) \] \[ =0.6020 \] Therefore \[ 32 = \frac{2.303\times0.6020}{k} \] \[ 32 = \frac{1.386}{k} \] \[ k=\frac{1.386}{32} \]

Step 3: Calculate half-life.
For first-order reaction, \[ t_{1/2} = \frac{0.693}{k} \] Substituting \(k=\frac{1.386}{32}\), \[ t_{1/2} = \frac{0.693\times32}{1.386} \] \[ =16\ \text{min} \]

Step 4: Final answer.
\[ {16\ \text{minutes}} \]